Chapter 16, Problem 80A

To determine

To find: The final water level in the glass beaker for the given temperatures.

Answer to Problem 80A

When a glass beaker is heated to $65°\text{C}$

  $V_2=601.85\text{ mL}$

When a glass beaker is cooled to $5°\text{C}$

  $V_3=46.26\text{ mL}$

Explanation of Solution

Given:

  $\begin{array}{c}{V}_1=250\text{mL}\\ T_1=27\text{C}\\ \text{=27+273}\\ \text{=300}\text{K}\\ T_2=65\text{C}\\ \text{=65+273}\\ \text{=338K}\end{array}$

Formula used:

Volume of water is directly proportional to the temperature at constant pressure.

  $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ .

  $\frac{V_3}{T_3}=\frac{V_2}{T_2}$ .

Calculation:

When a glass beaker is heated to $65°\text{C}$

  $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

  $V_2=\frac{250\text{ mL}}{27°\text{C}}\times 65°\text{C}$

  $V_2=601.85\text{ mL}$

The final volume of the water in the beaker is 601 .85 mL.

When a glass beaker is cooled to $5°\text{C}$

  $\begin{array}{c}{V}_2=601.85\text{ mL}\\ T_2=65°\text{C}\\ \text{=65+273}\\ \text{=338K}\\ T_3=5°\text{C}\\ \text{=5+273}\\ \text{=278K}\end{array}$

  $\frac{V_3}{T_3}=\frac{V_2}{T_2}$

  $V_3=\frac{V_2}{T_2}{T}_3$

  $V_3=\frac{601.85\text{ mL}}{65°\text{C}}\times 5°\text{C}$

  $V_3=46.26\text{ mL}$

The water in the beaker after the temperature decreased.